\newproblem{lay:7_4_18}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.4.18}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Show that if $A$ is square, then $|\det\{A\}|$ is the product of the singular values of $A$.
}{
   % Solution
	Let $A=U\Sigma V^T$ be a Singular Value Decomposition of the matrix $A$. If $A$ is square, then we may calculate its determinant and its absolute value as
	\begin{center}
		$|\det\{A\}|=|\det\{U\Sigma V^T\}|=|\det\{U\}\det\{\Sigma\}\det\{V^T\}|=|\det\{U\}||\det\{\Sigma\}||\det\{V^T\}|$
	\end{center}
	$U$ and $V$ are orthogonal matrices. This implies that their determinant is 1 or -1. Then,
	\begin{center}
		$|\det\{A\}|=|\det\{\Sigma\}|$
	\end{center}
	But $\Sigma$ is a diagonal matrix, so its determinant is the product of its diagonal entries, that are the singular values of $A$ (which are all non-negative values)
	\begin{center}
		$|\det\{A\}|=\left|\prod\limits_{i=1}^n{\Sigma_{ii}}\right|=\prod\limits_{i=1}^n{|\Sigma_{ii}|}=\prod\limits_{i=1}^n{\sigma_{i}}$
	\end{center}
}
\useproblem{lay:7_4_18}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

